Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct top {
  int a[10];
  top() { memset(a, 0x3f, sizeof(a)); }
  int operator[](int x) { return a[x]; }
  void add(int x) {
    if (x < a[9]) a[9] = x;
    for (int i = 8; i >= 0; i--)
      if (a[i + 1] < a[i]) swap(a[i + 1], a[i]);
  }
};
top merge(top a, top b) {
  top c;
  for (int i = 0; i < 10; i++) c.add(a[i]), c.add(b[i]);
  return c;
}
int anc[100001][17], d[100001];
top th[100001][17];
vector<int> g[100001];
top ans;
void dfs(int x, int p) {
  for (int y : g[x]) {
    if (y == p) continue;
    anc[y][0] = x;
    d[y] = d[x] + 1;
    dfs(y, x);
  }
}
int climb(int y, int num) {
  for (int i = 0; i < 17; i++)
    if ((num >> i) & 1) ans = merge(ans, th[y][i]), y = anc[y][i];
  return y;
}
int main() {
  int n, m, q, i, j, a, b, c;
  scanf("%d%d%d", &n, &m, &q);
  for (i = 0; i < n - 1; i++)
    scanf("%d%d", &a, &b), g[a].push_back(b), g[b].push_back(a);
  for (i = 1; i <= m; i++) scanf("%d", &a), th[a][0].add(i);
  anc[1][0] = 1;
  dfs(1, 0);
  for (i = 1; i < 17; i++)
    for (j = 1; j <= n; j++)
      anc[j][i] = anc[anc[j][i - 1]][i - 1],
      th[j][i] = merge(th[j][i - 1], th[anc[j][i - 1]][i - 1]);
  while (q--) {
    scanf("%d%d%d", &a, &b, &c);
    ans = top();
    if (d[a] > d[b])
      a = climb(a, d[a] - d[b]);
    else
      b = climb(b, d[b] - d[a]);
    for (i = 16; i >= 0; i--) {
      if (anc[a][i] == anc[b][i]) continue;
      ans = merge(ans, th[a][i]);
      ans = merge(ans, th[b][i]);
      a = anc[a][i];
      b = anc[b][i];
    }
    if (a != b) ans = merge(ans, merge(th[a][0], th[b][0])), a = anc[a][0];
    ans = merge(ans, th[a][0]);
    vector<int> pr;
    for (i = 0; i < 10; i++)
      if (ans[i] != 0x3f3f3f3f) pr.push_back(ans[i]);
    printf("%d ", min(c, (int)pr.size()));
    for (i = 0; i < min(c, (int)pr.size()); i++) printf("%d ", pr[i]);
    printf("\n");
  }
}